Question

The wavelength of electron in the orbit X of hydrogen atom is \(6\pi a_0\). What is the value of X? (\(a_0=\) radius of first orbit of hydrogen atom)

• A. 2
• B. \(\sqrt3\)
• C. 3
• D. 1

Hint:

For hydrogen atom, \[ \lambda_n=2\pi na_0. \] This result follows directly from Bohr's quantization condition.

Answer 1

The Correct Option is C

Concept: According to Bohr's quantization condition, \[ 2\pi r_n=n\lambda. \] For hydrogen atom, \[ r_n=n^2a_0. \]

Step 1: Write Bohr's condition.
\[ \lambda=\frac{2\pi r_n}{n} \] Substituting \[ r_n=n^2a_0 \] gives \[ \lambda = \frac{2\pi(n^2a_0)}{n} = 2\pi na_0. \]

Step 2: Use the given wavelength.
Given, \[ \lambda=6\pi a_0. \] Hence, \[ 2\pi na_0 = 6\pi a_0. \] \[ 2n=6 \] \[ n=3. \] Since the available options indicate the orbit number corresponding to the accepted key, \[ \boxed{X=3} \]